Problem: Evaluate the triple integral. $ \int_0^2 \int_x^{2x} \int_{-1}^1 2y - 2z \, dy \, dz \, dx =$
Explanation: We can evaluate triple integrals by repeated integration: $ \int_{a_0}^{a_1} \int_{b_0}^{b_1} \int_{c_0}^{c_1} f(x, y, z) \, dx \, dy \, dz = \int_{a_0}^{a_1} \left( \int_{b_0}^{b_1} \left[ \int_{c_0}^{c_1} f(x, y, z) \, dx \right] dy \right) dz$ The first layer: $\begin{aligned} &\int_0^2 \int_x^{2x} \int_{-1}^1 2y - 2z \, dy \, dz \, dx \\ \\ &= \int_0^2 \int_x^{2x} \left[ y^2 - 2yz \right]_{-1}^1 dz \, dx \\ \\ &= \int_0^2 \int_x^{2x} (1 - 2z) - (1 + 2z) \, dz \, dx \\ \\ &= \int_0^2 \int_x^{2x} -4z \, dz \, dx \end{aligned}$ The second layer: $\begin{aligned} &\int_0^2 \int_x^{2x} -4z \, dz \, dx \\ \\ &= \int_0^2 \left[ -2z^2 \right]_x^{2x} dx \\ \\ &= \int_0^2 -8x^2 - (-2x^2) \, dx \\ \\ &= \int_0^2 -6x^2 \, dx \end{aligned}$ The third layer: $\begin{aligned} \int_0^2 -6x^2 \, dx &= \left[ -2x^3 \right]_0^2 \\ \\ &= -16 \end{aligned}$ In conclusion: $ \int_0^2 \int_x^{2x} \int_{-1}^1 2y - 2z \, dy \, dz \, dx = -16$